In an electromagnetic wave in free space the root mean square value of the electric field is $E_{rms} = 6$ V/m

Question

In an electromagnetic wave in free space the root mean square value of the electric field is $E_{rms} = 6$ V/m. The peak value of the magnetic field is:

Accepted Answer

1. **Relationship between fields:** In an electromagnetic wave, the amplitudes of the electric field ($E_0$) and magnetic field ($B_0$) are related by the speed of light ($c$): $c = \frac{E_0}{B_0}$. Similarly, the rms values are related by $c = \frac{E_{rms}}{B_{rms}}$ .
2. **Calculate Peak Electric Field ($E_0$):** The peak value is related to the rms value by $E_0 = \sqrt{2} E_{rms}$.
   $$ E_0 = \sqrt{2} 	imes 6 	ext{ V/m} \approx 1.414 	imes 6 = 8.484 	ext{ V/m} $$
3. **Calculate Peak Magnetic Field ($B_0$):** Using $B_0 = \frac{E_0}{c}$ where $c = 3 	imes 10^8$ m/s .
   $$ B_0 = \frac{8.484}{3 	imes 10^8} = 2.828 	imes 10^{-8} 	ext{ T} $$
   Rounding to two decimal places, $B_0 \approx 2.83 	imes 10^{-8}$ T.
   Alternatively, find $B_{rms} = E_{rms}/c = 2 	imes 10^{-8}$ T, then $B_0 = \sqrt{2} B_{rms} = 1.414 	imes 2 	imes 10^{-8} \approx 2.83 	imes 10^{-8}$ T.

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